Question

in JAVA please thanks. (need answer in a few hours !!)

Exercise #1: Design and implement a program (name it LinearBinarySearch) to implement and test the linear and binary search algorithm discussed in the lecture slides. Define method LinearSearch() to implement linear search of an array of integers. Modify the algorithm implementation to count number of comparisons it takes to find a target value (if exist) in the array. Define method BinarySearch() to implement binary search of an array of integers. Modify the algorithm implementation to count number of comparisons it takes to find a target value (if exist) in the array. Now, develop a test method to read integer values from the user in to an array and then call methods Linear Search() and BinarySearch() and printout the number of comparison took to find the target values using each search method. Document your code and organized your outputs as follows: Arrays Values: Target value: Linear Search Comparisons: Binary Search Comparisons:

Answer 1

Program:

import java.util.Arrays;
import java.util.Scanner;

public class SearchAlgorithm {

   public static void main(String[] args) {
       int noOfComparisons;
   Scanner scannerObj=new Scanner(System.in); //scanner object creation - read user input
   System.out.println("Enter Array Length:");
   int size=scannerObj.nextInt(); //input Array size
int arr[] = new int[size];
int sortedArray[]=new int[size];
System.out.println("Enter Array Elements:");
for(int index=0;index<size;index++)
{
   arr[index]=scannerObj.nextInt(); //input Array Elements
}
System.out.println("Enter Element to Search:");
int key = scannerObj.nextInt(); //user input Element to Search
int last=arr.length-1;
noOfComparisons=linearSearch(arr,key); //calling linearsearch method
System.out.println("No of Comparisons in LinearSearch:"+noOfComparisons); //printing no of comparisons
Arrays.sort(arr); //For binary search - need to sort elements
for(int index=0;index<size;index++)
   sortedArray[index]=arr[index];
noOfComparisons= binarySearch(sortedArray,0,last,key); //calling binary search method
System.out.println("No of Comparisons in Binary Search:"+noOfComparisons);
  
}
   public static int binarySearch(int arr[], int first, int last, int key){
       int noOfComparisons=0;
       System.out.println("\t\t Binary Search- Array Elements: \t\t");
   for(int index=0;index<arr.length;index++)
       System.out.print(arr[index]+","); //display array elements
       int mid = (first + last)/2;
       while( first <= last ){
       if ( arr[mid] < key ){
       first = mid + 1; //binary search logic
       noOfComparisons++;
       }else if ( arr[mid] == key ){
       System.out.println("Element is found at index: " + mid);
       break;
       }else{
       last = mid - 1;
       }
       mid = (first + last)/2;
       }
       if ( first > last ){
       System.out.println("Element is not found!");
       }
       return noOfComparisons;
       }
     
   public static int linearSearch(int arr[],int key)
   { boolean elementStatus=false;
   int noOfComparisons=0;
   int index;
   System.out.println("\t\t Linear Search- Array Elements: \t\t");
   for(index=0;index<arr.length;index++)
       System.out.print(arr[index]+",");
       for(index=0;index<arr.length;index++)
       { noOfComparisons++; //linear search logic
           if(arr[index]==key)
           {
               elementStatus=true;
               break;
           }
       }
       if(elementStatus==true) {
           System.out.println("Element is found at index: " + index);
       }
       else
           System.out.println("Element is not found!");
       return noOfComparisons;
   }
}

Output:

Enter Array Length: Enter Array Elements: 18 Enter Element to Search: 17 Linear Search- Array Elements: 2,3,17,13,1,18, Element is found at index: 2 No of Comparisons in Linear Search: 3 Binary Search- Array Elements: 1,2,3, 13, 17, 18, Element is found at index: 4 No of Comparisons in Binary Search:1